Question

tìm x

x³-16x=0

3x(2-x)-2+x=0

(x+3)(x²-2x+3) = (x+3) (5-2x)

Answer 1

x³-16x=0

=> x(x²-16)=0

=> x(x-4)(x+4)=0

=> \(\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

vậy x=0 ;x=4;x=-4

3x(2-x)-2+x=0

=> 3x(2-x)-(2-x)=0

=> (2-x)(3x-1)=0

=> \(\left[{}\begin{matrix}2-x=0\\3x-1=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=2\\3x=1\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)

vậy x=2 hoặc x=\(\dfrac{1}{3}\)

Answer 2

c) (x+3)(x²-2x+3)=(x+3)(5-2x)

=>(x+3)(x²-2x+3) - (x+3)(5-2x)=0

=>(x+3)(x²-4x-2)=0

=>\(=>\left[{}\begin{matrix}x+3=0\\\text{x^2-4x-2}=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2=6\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=\sqrt{6}+2\\x=-\sqrt{6}+2\end{matrix}\right.\)

Answer 3

a)x³-16x=0

x(x²-16)=0

Hoặc x=0 hoặc x²-16=0

hoặc x=0 hoặc x=4

b)3x(2-x)-2+x=0

3x(2-x)-(2-x)=0

(2-x)(3x-1)=0

Hoặc 2-x=0 hoặc 3x-1=0

hoặc x=2 hoặc x=\(\dfrac{1}{3}\)

Câu c từ từ mik làm nha bạn